Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 1}{x + 10} = \dfrac{-8x + 19}{x + 10}$
Answer: Multiply both sides by $x + 10$ $ \dfrac{x^2 - 1}{x + 10} (x + 10) = \dfrac{-8x + 19}{x + 10} (x + 10)$ $ x^2 - 1 = -8x + 19$ Subtract $-8x + 19$ from both sides: $ x^2 - 1 - (-8x + 19) = -8x + 19 - (-8x + 19)$ $ x^2 - 1 + 8x - 19 = 0$ $ x^2 - 20 + 8x = 0$ Factor the expression: $ (x - 2)(x + 10) = 0$ Therefore $x = 2$ or $x = -10$ At $x = -10$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -10$, it is an extraneous solution.